Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+4y &= 1 \\ -8x+8y &= -6\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $8y = 8x-6$ Divide both sides by $8$ to isolate $y$ $y = {x - \dfrac{3}{4}}$ Substitute this expression for $y$ in the first equation. $8x+4({x - \dfrac{3}{4}}) = 1$ $8x + 4x - 3 = 1$ Simplify by combining terms, then solve for $x$ $12x - 3 = 1$ $12x = 4$ $x = \dfrac{1}{3}$ Substitute $\dfrac{1}{3}$ for $x$ back into the top equation. $8( \dfrac{1}{3})+4y = 1$ $\dfrac{8}{3}+4y = 1$ $4y = -\dfrac{5}{3}$ $y = -\dfrac{5}{12}$ The solution is $\enspace x = \dfrac{1}{3}, \enspace y = -\dfrac{5}{12}$.